Subject: Chemistry — Chemical Equilibrium
Question: Consider the following reversible reaction,
$$A(g) + B(g) \rightleftharpoons AB(g)$$
The activation energy of the backward reaction exceeds that of the
forward reaction by $2\ RT$ (in $J\ mol^{- 1}$).
If the pre-exponential factor of the forward reaction is 4 times that
of the reverse reaction, the absolute value of
$\mathrm{\Delta}G{^\circ}$ (in $J\ mole^{- 1}$) for the reaction at
300 K is \_\_\_\_\_.
(Given: $\ln(2) = 0.7,\ RT = 2500\ J\ mol^{- 1}$ at $300\ K$ and $G$
is the Gibbs energy)
Answer: 8500 J/mole
Solution: $A(g) + B(g) \rightleftharpoons AB(g)$
$E_{a_{b}} = E_{a_{f}} + 2RT$ & $A_{f} = 4A_{b}$
Now, Rate constant of forward reaction $k_{f} = A_{f}e^{- Ea_{f}/RT}$
Rate constant of reverse reaction $K_{b} = A_{b}e^{E_{ab}/RT}$
Equilibrium constant
$$K_{eq} = \frac{K_{f}}{K_{b}} = \frac{A_{f}}{A_{b}}e^{- \left( E_{af} - E_{ab} \right)/RT}$$
$$= 4e^{+ 2} = 4e^{2}$$
Now,
$\mathrm{\Delta}G{^\circ} = - RT\ln K_{eq} = - 2500\ln\left( 4e^{2} \right)$
$= - 2500\left( \ln 4 + \ln e^{2} \right)$
$= - 2500(1.4 + 2) = - 2500 \times 3.4$
$= 8500\ J/mole$
"Absolute value $= 8500\ J/mole$"